Tính \(\mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{{x^2} + 6x - 8}}{{{x^2} - 4}}\) bằng:

* Hướng dẫn giải

Ta có

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} + 6x - 8} \right) = 8 > 0\\\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 0;\,\,x \to {2^ + } \Rightarrow x - 2 > 0\\ \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{{x^2} + 6x - 8}}{{{x^2} - 4}} =  + \infty \end{array}\)

Chọn A.